Electric field due to a uniformly charged disc

AI Thread Summary
The discussion focuses on deriving the electric field due to a uniformly charged disc, specifically the integration process involved in calculating the total electric field from an elementary ring. The initial formula for the electric field contribution from the ring is provided, but the user expresses confusion about the calculus needed to complete the derivation. Suggestions include integrating both sides and using trigonometric substitution to simplify the integral. Participants discuss the formatting issues with TeX in the forum, which affects the clarity of the mathematical expressions. The conversation concludes with the user deciding to apply the suggested method for solving the integral.
vinzie
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At the end of the derivation, it is given
The electric fiel due to elementary ring at the point P is dE = [2∏rσdrx]/[4∏epsilon zero (x^2 +r^2)^(3/2)
]

To find the total E due to disc is given by

∏σx/4∏ε(2rdr)/(x2 + r2)3/2

I am stuck with the calculus done here to arrive at the solution. Please help me.

Thank You!

Vinzie
 
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Do you mean

\mathrm{d}E=\dfrac{2\cdot\pi\cdot r\cdot\sigma\cdot x\cdot\mathrm{d}r}{4\cdot\pi\cdot\epsilon_0\cdot\left(x^2+r^2\right)^{\frac32}}

and you want to know how to determine that

E=\dfrac{\pi\cdot\sigma\cdot x}{4\cdot\pi\cdot\epsilon_0}\cdot\displaystyle\int\left(\dfrac{2\cdot r\cdot\mathrm{d}r}{\left(x^2+r^2\right)^{\frac32}}\right)

Also, why isn't the first TeX parsing?

Simple. Integrate both sides.
 
I can't see your formula!
 
Yep, for some reason, the TeX isn't parsing. It should! It does on AoPS!
 
Whovian said:
Do you mean

\mathrm{d}E=\dfrac{2\cdot\pi\cdot r\cdot\sigma\cdot x\cdot\mathrm{d}r}{4\cdot\pi\cdot\epsilon_0\cdot\left(x^2+r^2\right)^{\frac32}}

and you want to know how to determine that

E=\dfrac{\pi\cdot\sigma\cdot x}{4\cdot\pi\cdot\epsilon_0}\cdot\displaystyle\int\left(\dfrac{2\cdot r\cdot\mathrm{d}r}{\left(x^2+r^2\right)^{\frac32}}\right)

Also, why isn't the first TeX parsing?

Simple. Integrate both sides.

Hi,

WOuld you mind if you tell me how to solve the integration part?

Thank You
 
Divid (x^2+r^2) \; by r^2\; to get 1+(\frac x r)^2\; and use trig function to substitute. and see what happen. Something like \tan \theta=\frac x r. 1+\tan^2\theta=\sec^2\theta
 
Last edited:
yungman said:
Divid (x^2+r^2) \; by r^2\; to get 1+(\frac x r)^2\; and use trig function to substitute. and see what happen. Something like \tan \theta=\frac x r. 1+\tan^2\theta=\sec^2\theta

Thanks Yungman!

I am going to solve that way.
 
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